Math, asked by eratzinfantry, 1 month ago

If sin x is equal to 3/5 and X belongs to [π • 3π/ 2] then tan x is equal to​

Answers

Answered by shrabani1463
2

Answer:

DEF are respectively the midpoints of sides AB, BC and CA of triangle ABC, then: Ratio of area of triangle DEF : area of triangle ABC = 1 : 4

Solution:

In the given question, we know Triangle DEF formed with midpoints is similar to the Outer Triangle ABC

On the basis of the similarity, we can say,

If two triangles are similar then the ratio of their area is equal to the square of the ratio of their corresponding sides

Mathematically can be written as :-

\frac{\text {area of triangle } D E F}{\text {area of triangle ABC}}=\left(\frac{D E}{A C}\right)^{2}

area of triangle ABC

area of triangle DEF

=(

AC

DE

)

2

Since, DECF is a parallelogram. So DE = FC

On substituting:

\frac{\text {area of triangle } D E F}{\text {area of triangle ABC}}=\left(\frac{F C}{A C}\right)^{2}

area of triangle ABC

area of triangle DEF

=(

AC

FC

)

2

Also, F is the midpoint of AC

\text { So } \mathrm{AC}=2 \times \mathrm{FC} So AC=2×FC

\begin{gathered}\begin{array}{l}{\frac{\text {area of triangle } D E F}{\text {area of triangle } A B C}=\left(\frac{F C}{2 F C}\right)^{2}} \\\\ {\frac{\text {area of triangle } D E F}{\text {area of triangle } A B C}=\frac{1}{4}}\end{array}\end{gathered}

area of triangle ABC

area of triangle DEF

=(

2FC

FC

)

2

area of triangle ABC

area of triangle DEF

=

4

1

Hence ratio of area of triangle DEF and triangle ABC is given as:

Ratio of area of triangle DEF : area of triangle ABC = 1 : 4

Learn more about midpoints of triangle

Find the vertices of a triangle, the midpoints of whose side are(3,1),(5,6,),(-3,2)

https://brainly.in/question/3073037

The mid-points of the sides of a triangle ABC along with any of the vertices as the fourth point make a parallelogram of area equal to what?

https://brainly.in/question/5968686

Answered by arvindkurdekar2
0

Answer:

Given as sin x = 3/5, tan y = 1/2 and π/2 < x< π< y< 3π/2 As we know that, x is in second quadrant and y is in third quadrant. Since, in second quadrant, cos x and tan x are negative. Since, in third quadrant, sec y is negative. On using the formula, cos x = – √(1 - sin2 x) tan x = sin x/cos x Then, cos x = – √(1 - sin2 x) = – √(1 – (3/5)2) = – √(1 – 9/25) = – √((25 - 9)/25) = – √(16/25) = – 4/5 tan x = sin x/cos x = (3/5)/(-4/5) = 3/5 × -5/4 = -3/4 As we know that sec y = – √(1 + tan2 y) = – √(1 + (1/2)2) = – √(1 + 1/4) = – √((4 + 1)/4) = – √(5/4) = – √5/2 Then, 8 tan x – √5 sec y = 8(-3/4) – √5(-√5/2) = -6 + 5/2 = (-12 + 5)/2 = -7/2 Thus, 8 tan x – √5 sec y = -7/2Read more on Sarthaks.com - https://www.sarthaks.com/652839/if-sin-x-3-5-tan-y-1-2-and-2-x-y-3-2-find-the-value-of-8-tan-x-5-sec-y

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